For what value of $c$ will the circle with equation $x^2 + 6x + y^2 - 4y + c = 0$ have a radius of length 4?
Answer: Completing the square gives us $(x+3)^2 + (y-2)^2 = 13 - c$. Since we want the radius to be 4, we must have $13 - c = 4^2$. It follows that $c = \boxed{-3}$.